package programmercarl.动态规划.C48;

class Solution1 {
    public int numDistinct(String s, String t) {
        //dp含义是以i-1和j-1为结尾的s和t的的不同子序列的个数
        int m= s.length();
        int n = t.length();
        int[][] dp = new int[m+1][n+1];
        for (int i = 0; i <= m; i++) {
            dp[i][0]=1;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <=n ; j++) {
                if (s.charAt(i-1) == t.charAt(j-1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[m][n];
    }
}